What is the integral from 1 to 2 of 2x^2sqrt(x^3+1) ?

The integral from 1 to 2 of 2x^2sqrt(x^3+1) is (2(54-4sqrt(2)))/9

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integral from 1 to 2 of 2x^2sqrt(x^3+1)

\int_{1}^{2} 2 x^{2} x^{3} + 1 d x

\frac{2 ( 54 - 4 2 )}{9}

Decimal Notation

10.74292 \dots

Solution steps

\int_{1}^{2} 2 x^{2} x^{3} + 1 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{2} x^{2} x^{3} + 1 d x

Apply u-substitution

: \int_{2}^{9} \frac{u}{3} d u

= 2 \cdot \int_{2}^{9} \frac{u}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{3} \cdot \int_{2}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9}

= 2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9}

Simplify

2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9} : \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9}

= \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9}

Compute the boundaries:

18 - \frac{4 2}{3}

= \frac{2}{3} (18 - \frac{4 2}{3})

Simplify

= \frac{2 ( 54 - 4 2 )}{9}

\frac{d}{d x} (\frac{e ^{x}}{sin ( 2 x )})

\int x (2 x^{2} + 3)^{4} d x

x \to 0 lim (\frac{sin ( x )}{- x})

\frac{d}{d y} (x^{\frac{y}{z}})

- sin (x^{2})

What is the integral from 1 to 2 of 2x^2sqrt(x^3+1) ?
The integral from 1 to 2 of 2x^2sqrt(x^3+1) is (2(54-4sqrt(2)))/9